Eigenfunction of ergodic skew product fixed by commutator?

Question

Background: Let $(Y, \mathcal{B},\mu,T)$ be an ergodic probability system and let $G$ be a compact metrizable group with compact subgroup $H$. Given a measurable map $\rho:Y \to G$. We may define the skew product system $W_{H,\rho}=(Y \times G/H, \mu \times m_{G/H}, T_{\rho})$ where the transformation is given by $$ T_{\rho}(y,u)=(Ty, \rho(y)u)$$ and by $m_{G/H}$ we mean the pushforward onto $G/H$ of the Haar measure on $G$. Note that for $g \in G$ we have a measure preserving map $g:Y \times G/H \to Y \times G/H$ given by $$g.(y,u)=(y,gu)$$ and thus $G$ acts on the Hilbert Space $L^2(W_{H,\rho})$.

Question: If the system $(Y \times G, \mu \times m_G, T_{\rho})$ is ergodic, then is it true that for any eigenfunction $e: Y \times G/H \to \mathbb{C}$ of $W_{H,\rho}=(Y \times G/H, \mu \times m_{G/H}, T_{\rho})$ we have that $e \circ [g_1,g_2]=e$ for $g_1,g_2 \in G$? i.e all eigenfunctions are fixed by the commutator subgroup $[G,G]$?

Motivation: It seems to me that this is suggested in the proof of Lemma 6.1 in Nonconventional ergodic averages and nilmanifolds by Host-Kra (available to public here http://www.math.northwestern.edu/~kra/papers/convnil.pdf)

Answer 1

Yes, the eigenfunctions of $X = Y\times G/H$ come from $Y\times A$, where $A$ is an abelian quotient of $G$. I'll indicate how to construct $A$, checking the required properties is then routine. Note that we may assume $H$ trivial, since eigenfunctions of $Y\times G/H$ are also eigenfunctions of $Y\times G$.

Note that the $Y$-module spanned by the eigenfunctions is an algebra, and since it is spanned by rank 1 submodules it corresponds to an abelian group extension $Y\times A$ with a minimal cocycle $\alpha$. The trick is now to relate $G$ and $A$. To do this note that we have a measure-theoretic factor map $X \to Y\times (G\times A)$, where the latter system is equipped with the cocycle $\rho\times\alpha$ and the measure on the target system is a pushforward measure.

By the Mackey group construction we may assume that $\rho\times\alpha$ is minimal with values is a subgroup $K\leq G\times A$. Since $\rho$ and $\alpha$ are minimal, $K$ has full projections on both coordinates. Moreover, the projection map $Y\times K \to Y\times G$ is a measure-theoretic isomorphism, so the coordinate projection $K\to G$ is in fact a compact group isomorphism. This gives the desired morphism $G\to K\to A$.